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NEET levelSingle correct optionDifficulty 1 of 3mc059

$2.4 g$ of $Mg$ ( $24 g/mol$ ) is added to a solution containing $0.15 mol$ of $HCl$ according to $Mg + 2 HCl MgCl X_{2} + H X_{2}$ . The volume of $H X_{2}$ at STP is

A $2.24 L$
B $3.36 L$
C $5.60 L$
D $1.68 L$

Answer: D

Mg

= 0.1 mol needs 0.2 mol

HCl

, but only 0.15 mol is present, so

HCl

is limiting.

H X_{2}

= 0.15/2 = 0.075 mol. Volume = 0.075 x 22.4 = 1.68 L.

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